Finding the sum of the first $n$ terms of an AP

For a finite arithmetic progression, the governing formula is S_n = n/2(a + l), where a is the first term and l is the nth, or last, term. Here a = 9, d = 13 − 9 = 4, and n = 15. First calculate the fifteenth term: l = a + (n − 1)d = 9 + 14 × 4 = 65. Therefore S_15 = 15/2(9 + 65) = 15/2 × 74 = 15 × 37 = 555. Thus option A is correct. The alternate formula confirms this: S_15 = 15/2[2(9) + 14(4)] = 15/2(18 + 56) = 555. The nearby choices are distractors caused by an incorrect last term, an incorrect common difference, or an arithmetic slip.

This is the sum of the first (9) multiples of (4), so there are (180) plants. Even in real situations, write (a), (d), and (n).

For an arithmetic progression, the sum of n terms can be found using Sₙ = n/2 × (first term + last term). Substituting n = 13, the first term 2, and the last term 50 gives S₁₃ = 13/2 × (2 + 50) = 13/2 × 52 = 13 × 26 = 338. Therefore, option B is correct; the other values result from an arithmetic or formula error.

The first term is (100), and the eighth term is (30), so (S_8=\frac{8}{2}(130)=520). The same formula applies to decreasing progressions.

Direct answer: option C, 144. The odd natural numbers begin 1, 3, 5, 7, and continue by adding 2. For the first n odd numbers, the sum is n^2. Here n=12, so S=12^2=12×12=144. Equivalently, the terms are 1,3,5,7,9,11,13,15,17,19,21,23; pairing or adding them also gives 144. Option A, 121, is 11^2, so it would be the sum of the first 11 odd numbers, not 12. Option B, 132, does not follow the odd-number sum rule and is too small. Option C, 144, exactly matches 12^2, so it is correct. Option D, 156, is also not the required square and is too large. Remember: the sum of the first n odd numbers is always n^2.

The sequence is an arithmetic progression with first term a = 14 and common difference d = 17 - 14 = 3. To sum its first 10 terms, use S_n = n/2[2a + (n - 1)d]. Substitution gives S_10 = 10/2[2(14) + 9(3)] = 5[28 + 27] = 5 × 55 = 275. The result can also be checked by finding the tenth term: l = 14 + 9 × 3 = 41, followed by S_10 = 10/2(14 + 41) = 5 × 55 = 275. Hence option B is correct. The other options are close distractors caused by using the wrong number of differences, finding an incorrect last term, or making an arithmetic error.

The sum of the first \(n\) terms of an AP is the number of terms multiplied by the average of the first and last terms. Hence, \(S_n=n\times\frac{a+l}{2}=\frac{n}{2}(a+l)\). Option B does not divide by \(2\), while option C gives only the average of the first and last terms, not the sum. Exam tip: use this formula directly when the last term \(l\) is known.

The direct answer is option B, 270. The first term is 50 and the common difference is 45−50=−5. The ninth term is a9=a1+8d=50+8(−5)=50−40=10. Now apply the sum formula: S9=9/2(a1+a9)=9/2(50+10)=9/2×60=270. Equivalently, the nine terms are 50,45,40,35,30,25,20,15,10, whose sum is 270. Option A, 260, is 10 too small. Option B, 270, is correct. Option C, 280, is 10 too large, and option D, 290, is even larger. The negative common difference means the terms decrease, but the sum formula remains unchanged. Remember to use n−1=8 differences when finding the ninth term, not 9 differences.

The sum of the (7)th to (12)th terms is (S_{12}-S_6=135). Find the sum of middle terms by subtracting partial sums.

This is the sum of the first (8) multiples of (7), so (7\times36=252). For multiples, use the sum of natural numbers.

The sum of the first \(n\) terms of an AP equals the number of terms multiplied by the average of the first and last terms. Thus, \(S_n=n\times\frac{a+l}{2}=\frac{n}{2}(a+l)\). Therefore, option B is correct. Option A misses division by 2, while option C gives only the average of the first and last terms. Exam tip: use this formula directly when the last term \(l\) is known.

Direct answer: option C, 301. In the AP 2,5,8,..., the first term is a=2 and the common difference is d=3. The fourteenth term is a14=a+(14-1)d=2+13×3=41. The AP sum formula is S_n=n/2[2a+(n-1)d], or n/2(first term+last term). Thus S14=14/2×(2+41)=7×43=301. Option A, 296, results from an incorrect final term or arithmetic and is not the sum. Option B, 299, is also inconsistent with the formula. Option C, 301, uses the correct fourteenth term and is correct. Option D, 303, is slightly too large and does not result from the AP calculation. A useful check is that the average of the first and last terms is (2+41)/2=21.5; multiplying by 14 gives 301.

Here (a=100), (d=50), and (n=6), so the total saving is (1350) rupees. In money questions, keep the unit in mind.

The ninth term is (40), so (S_9=\frac{9}{2}(16+40)=252). The average of the first and last terms is the middle term.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Substituting the given values, \(S_7=\frac{7}{2}(25+5)=\frac{7}{2}\times30=105\). Therefore, the correct answer is 105. Option 110 is incorrect because \(\frac{7}{2}\) times 30 equals 105, not 110. Exam tip: first add the values inside the brackets, then simplify with 2 before multiplying whenever possible.

This is the sum of the first (5) multiples of (11), so (11\times15=165). For multiples, the sum of the first (n) natural numbers is useful.

The sum of the first 10 terms includes the sum of the first 5 terms. Therefore, the sum of the 6th to 10th terms is 145 - 45 = 100. Obtaining 95 would result from an incorrect subtraction. Exam tip: To find the sum of a consecutive block of terms, subtract the smaller partial sum from the larger partial sum.

Answer: 252, so option B is correct. The first term is a = 18, the common difference is d = 24 - 18 = 6, and the number of terms is n = 7. First find the seventh term using a_n = a + (n - 1)d: a_7 = 18 + (7 - 1)6 = 18 + 36 = 54. Now use the sum formula S_n = n(a + l)/2, where l is the last term. Thus S_7 = 7(18 + 54)/2 = 7×72/2 = 7×36 = 252. We can also list the terms: 18, 24, 30, 36, 42, 48, 54; their total is 252. Option A, 250, option C, 254, and option D, 256, are close numerical distractors and do not equal the calculated sum. A useful check is that the average of the first and last terms is (18+54)/2=36, and 7×36=252.

Using \(\text{average}=\frac{\text{sum}}{\text{number of terms}}\), the sum is \(18\times 8=144\). Therefore, 144 is the correct option. A value such as 148 would result from using an incorrect average or number of terms. Exam tip: When the average and the number of terms are given, multiply them to find the sum.

The tenth term is (13), so (S_{10}=\frac{10}{2}(40+13)=265). In a decreasing progression, the last term is smaller.

This is the sum of the first (50) natural numbers, which is (1275). When (d=1) and (a=1), use the natural-number sum formula.

These are the (3)rd to (8)th multiples of (7), or directly (a=21), (d=7), (n=6) gives (231). Look carefully at the first term.

The eighth term is (59), so (S_8=\frac{8}{2}(17+59)=304). Finding the last term first makes the sum easier.

Here (a=12), (d=4), and (n=11), so there will be (352) books. In word problems, treat the given pattern as an arithmetic progression.

Here, the first term is \(a=6\), the common difference is \(d=9-6=3\), and the number of terms is \(n=14\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{14}=\frac{14}{2}[2(6)+13(3)]=7(51)=357\). Hence, 357 is correct. An option such as 367 usually results from an error while calculating \((n-1)d\) or adding the terms. Exam tip: write down \(a\), \(d\), and \(n\) separately before applying the sum formula.